cp's OEIS Frontend

A160364 Let f be defined as in A159885 and f^k be the k-th iteration of f. Then a(n) is the least k for which either {A000120(f^k(2n+1)) < A000120(2n+1)}&{A006694((f^k(2n+1)-1)/2)<=A006694(n)} or {A000120(f^k(2n+1))<=A000120(2n+1)}&{A006694((f^k(2n+1)-1)/2) < A006694(n)}.

%I A160364 #4 Jun 02 2025 01:39:56
%S A160364 2,1,1,5,3,1,1,2,5,1,2,1,1,1,1,5,2,5,3,33,3,1,1,1,1,1,1,1,1,3,1,5,7,1,
%T A160364 5,10,1,1,2,5,5,1,1
%N A160364 Let f be defined as in A159885 and f^k be the k-th iteration of f. Then a(n) is the least k for which either {A000120(f^k(2n+1)) < A000120(2n+1)}&{A006694((f^k(2n+1)-1)/2)<=A006694(n)} or {A000120(f^k(2n+1))<=A000120(2n+1)}&{A006694((f^k(2n+1)-1)/2) < A006694(n)}.
%C A160364 Using induction, one can prove that the Collatz (3x+1)-conjecture follows from the finiteness of a(n) for every n.
%e A160364 Beginning with n=1, we have f(2n+1)=f(3)=5. Here A000120(3)=A000120(5)=2 and A006694((3-1)/2)= A006694((5-1)/2)=1. None of values did not become less than. Therefore a(1)>1. Since f(5)=1 and A000120(1)=1 and A006694(0)=0, then a(2)=2.
%Y A160364 A000120 A006694 A160267 A006694 A122458 A160266 A159885 A159945 A160198
%K A160364 nonn,uned
%O A160364 1,1
%A A160364 _Vladimir Shevelev_, May 11 2009